# Find the equation of the line with slope -1 that is tangent to the curve y=1/x-1

by gibguitar
Tags: curve, equation, line, slope, tangent, y1 or x1
 P: 9 1. The problem statement, all variables and given/known data Find the equation of the line with slope -1 that is tangent to the curve y=1/x-1 2. Relevant equations y=1/x-1 3. The attempt at a solution Slope of -1 means y=-1x+k So... -1x+k = 1/x-1 I don't know how to rearrange this into a quadratic equation so that I can solve this. I know I use b^2-4ac=0 and substituting for a, b and c.... I'm just stuck on rearranging the equation into the form: ax^2+bx+c=0
Mentor
P: 21,397
 Quote by gibguitar 1. The problem statement, all variables and given/known data Find the equation of the line with slope -1 that is tangent to the curve y=1/x-1 2. Relevant equations y=1/x-1 3. The attempt at a solution Slope of -1 means y=-1x+k So... -1x+k = 1/x-1 I don't know how to rearrange this into a quadratic equation so that I can solve this. I know I use b^2-4ac=0 and substituting for a, b and c.... I'm just stuck on rearranging the equation into the form: ax^2+bx+c=0
You're missing a very important piece: how to find the slope of the tangent line to y = 1/x - 1.

I'm going to guess that you are in a class that has discussed how to find the tangent to a curve.
 Mentor P: 21,397 OTOH, maybe this actually is a precalculus-type problem. What does it mean to say that a line is tangent to a curve? For your equation, -x + k = 1/x - 1, what about multiplying both sides by x?
P: 9
Find the equation of the line with slope -1 that is tangent to the curve y=1/x-1

 Quote by Mark44 OTOH, maybe this actually is a precalculus-type problem. What does it mean to say that a line is tangent to a curve? For your equation, -x + k = 1/x - 1, what about multiplying both sides by x?
(-x+k)(x) = -x^2+k(x)
(1/x-1)(x) = ?

You're right, but I'm confused as to how to multiply the second part? x/x^2-x ?
 P: 9 Here's my second stab at it... -1x+k = 1/x-1 Rearranged: -x^2+kx-1 Using the discriminant of the quadratic formula: (b^2-4ac) a=-1 b=k^2 c=-1 k^2-4(-1)(-1)=0 Solving: k = -2 Therefore the equation is y=-1x-2 For the line with a slope of -1 that is tangent to the curve of 1/x-1 Can anybody verify if I got this right?
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P: 21,397
 Quote by gibguitar Here's my second stab at it... -1x+k = 1/x-1 Rearranged: -x^2+kx-1
Where did the = go? Also, you have an error.
 Quote by gibguitar Using the discriminant of the quadratic formula: (b^2-4ac) a=-1 b=k^2 c=-1 k^2-4(-1)(-1) Solving: k = -2 Therefore the equation is y=-1x-2 For the line with a slope of -1 that is tangent to the curve of 1/x-1 Can anybody verify if I got this right?
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P: 21,397
 Quote by gibguitar (-x+k)(x) = -x^2+k(x) (1/x-1)(x) = ? You're right, but I'm confused as to how to multiply the second part? x/x^2-x ?
What is x * (1/x)?
What is x * (-1)?
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P: 21,397
From post #3:
 What does it mean to say that a line is tangent to a curve?
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Thanks
P: 5,197
 Quote by gibguitar 1. The problem statement, all variables and given/known data Find the equation of the line with slope -1 that is tangent to the curve y=1/x-1 2. Relevant equations y=1/x-1 3. The attempt at a solution Slope of -1 means y=-1x+k So... -1x+k = 1/x-1 I don't know how to rearrange this into a quadratic equation so that I can solve this. I know I use b^2-4ac=0 and substituting for a, b and c.... I'm just stuck on rearranging the equation into the form: ax^2+bx+c=0
What you wrote means $(1/x) - 1.$ Is that what you meant, or did you really mean $1/(x-1)$? If you meant the latter, you need to use brackets, but if you meant the former then what you wrote is OK.

RGV
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P: 21,397
 Quote by Ray Vickson What you wrote means $(1/x) - 1.$ Is that what you meant, or did you really mean $1/(x-1)$? If you meant the latter, you need to use brackets, but if you meant the former then what you wrote is OK.
I think he meant (1/x) - 1, but I'm not certain of it.

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